MC ANSWERS

[stonecold]

This is what I got.  Not saying it's right, just for comaparison.  Let me know if you have something different or I'm wrong and I shall change it.

1. C
2. D
3. A
4. C
5. C
6. A
7. A
8. B
9. B
10. C
11. A
12. D
13. B
14. D
15. B
16. C
17. D
18. B
19. C
20. B

[qshyrn]

same, except i got c for 4 and d for 5   (what were these questions again?

[10weid]

C
D
A
C
C
A
A
B
B
C
A
D
B
D
B
C
D
B
C
B

[Studyinghard]

I only remember the first 5. Did anyone else think the first MC that all were wrong answers.

Not sure but was question 4 answer 2/7 mol? thats what I got.

and 5 was like 249L?

[stonecold]

nah, it was 5/7 mol, wasn't it in excess?

[matty.k]

i also got C for 4 5

[stonecold]

4 and 5 five are C c sorry, i mustve copied out wrong in a rush, becasue I got 249 L

[Studyinghard]

nah, it was 5/7 mol, wasn't it in excess?

yeah but when your doing ratios you dont take the difference in the mole. you do from the equation. which was 2/7

matty.k i got some answers for those questions as you.

[10weid]

yeah, 2/7 mol reacts, maening 5/7 mol is in excess, which was option C....

[Studyinghard]

Was 12 option D the one where which would react with dilute HCL and NaOH ? I put it the acid which was option C for the whole exam then with 2 mins left I changed it to the amino acid.

is 12 D for sure?

[10weid]

it was d, as an amino acid may act as an acod pr a abse, which is the answer rewquired

[Studyinghard]

phewww.

OO what did people do for the question about the concentration of oxalic acid. I did 0.750M

[10weid]

yeah, im farily certain it was .75, as the acid was clearly diprotic, meaning thta f the reacitonn goes to completion it gives away both protons, reuqiring 2 mol of NaOH

[Studyinghard]

winnnn...i changed that at the last second too XD

[kyzoo]

Hmm...apparently with the insulin one the number of peptide links is 49 because insulin is made of 2 strands??? Can someone confirm this? Did the question actually say insulin was made of 2 strands because I don't remember seeing that.